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Faraday's Law Problem

  1. Mar 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field away from vertical increases from 0.50 T to 1.50 T in 0.60 s. What is the induced emf (in mV) in the coil?

    2. Relevant equations

    [tex]\Phi_m = ABcos\vartheta[/tex]

    [tex]E = \frac{d \Phi}{dt}[/tex]

    3. The attempt at a solution

    [tex]E = \frac{d \Phi}{dt} = \frac{d (ABcos\vartheta)}{dt} = - \pi r^2 \frac{dB}{dt}sin\vartheta[/tex]

    [tex]\frac{dB}{dt} = \frac{1}{.6} = 1.67 T[/tex]

    I can "ignore" the negative sign because I just need the absolute value, and the derivative of cos = -sin

    *I forgot the number ot turns in the previous equation, but I added them in the next one*

    [tex]N \pi r^2 \frac{dB}{dt}sin\vartheta = (100) \pi 0.1^2 (1.67) sin60 = 45mV[/tex]

    Did I make any mistakes?
     
  2. jcsd
  3. Mar 25, 2008 #2

    G01

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    Your work looks correct except for the following point. (It doesn't affect your answer.)

    The minus sign you ignore should cancel out anyway since:

    [tex]E=-\frac{d\Phi}{dt}[/tex]

    You forgot the minus sign that comes from Lenz's Law.
     
  4. Mar 25, 2008 #3
    in my textbook, Farady's Law is denoted in an absolute value.

    apperantly I made a calculation error. I get 4.5 V = 4.5*10^-3 mV.

    EDIT: I tried 45 and 4.5*10^-3, and it says I got it wrong. Any ideas?
     
  5. Mar 25, 2008 #4

    G01

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    I also get 4.5 V when I crunch the numbers. But 4.5V is 4500mV, not .0045mV.
     
  6. Mar 25, 2008 #5
    I tried 4500 mV, but it still says it's wrong.
     
  7. Mar 25, 2008 #6

    G01

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    OK I think I see the problem. Your radius is.01m not .1m. Does this fix the problem?
     
  8. Mar 25, 2008 #7
    Firstly, i don't understand why you've taken the angle as [itex]\frac{\pi}{3}[/itex]. Since the field is vertical and the coil is in the horizontal plane, the angle is [itex]0[/itex] i.e. the area vector and the field lines are parallel. Secondly, the angle does not change with time. Hence, the term [itex]\cos(\theta)[/itex] is constant and ur mistake was taking it's derivative w.r.t time and changing it to [itex]\sin(\theta)[/itex]. My answer is coming out to be: 52.4645 mV [http://tinyurl.com/yv9w2k]
     
  9. Mar 25, 2008 #8
    I'm an idiot. Should have realised that.

    but why did you use 0.02 instead of 0.01 since it requires the radius and not the diameter.
     
  10. Mar 25, 2008 #9
    do note that i have taken the no. of turns as '25' and not '100' i.e. i used [tex]A = \frac{\pi d^2}{4}[/tex] rather than [itex]A = \pi r^2[/itex]
     
  11. Mar 25, 2008 #10
    shouldnt it be 25*pi*((0.02)^2)*(1.67)*cos60 = 26.23mV?
     
    Last edited: Mar 25, 2008
  12. Mar 25, 2008 #11

    G01

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    Hmm. Must have misread the problem. I was under the assumption that the OP just forgot to post the angle. in the problem. I figured that his value of 60 was correct. Sorry for the confusion.
     
  13. Mar 25, 2008 #12
    My idioticy has no bounds. The 30 degrees is from a different problem I'm working on and got those 2 mixed up.
     
  14. Mar 25, 2008 #13

    G01

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    Don't be so hard on yourself. It happens to everyone at some point or another. :smile:
     
  15. Mar 25, 2008 #14
    does the question say so? If it does.. then yes. Because in the problem you presented, no mention of angle is given. The question presents that the plane of the coil and the field lines of the magnetic field are perpendicular. So, i took the angle to be 0.
     
  16. Mar 25, 2008 #15
    Yes, but I already did 5/6 tries on the answer, and I have 1 more before I get a 0 on this question.

    Here's the full question:

    A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60 degrees away from vertical increases from 0.50 T to 1.50 T in 0.60 s.

    So would this be correct 25*pi*((0.02)^2)*(1.67)*cos60 = 26.23mV?
     
  17. Mar 25, 2008 #16
    well.. if that is the full question then this should be the right answer. Is this what your textbook/assignment/whatever says?
     
  18. Mar 25, 2008 #17
    this is what it says:

    [​IMG]
     
  19. Mar 25, 2008 #18
    i was actually asking what the textbook/assignment had to say about the 'answer'.. i didn't realize it then that this was a test you were giving. Anyways, all things considered, i think you should go with 23.24 or whatever the answer was that u got in ur previous post wherein you used cos(60).
     
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